The operating temperature of an indirectly heated filament of a vaccum tube is around 1050K. At what wavelength will it radiate maximum ? Given b = 0.288 cm K.

Explanation

To find the wavelength at which the filament radiates maximum intensity, we will use Wien's Displacement Law. This law states that the wavelength of maximum radiation (λ_max) is inversely proportional to the absolute temperature (T) of the body. The relationship can be expressed as:

\[ \lambda_{max} \cdot T = b \]

Where:
- \( \lambda_{max} \) is the wavelength at which the radiation is maximum,
- \( T \) is the absolute temperature in Kelvin,
- \( b \) is Wien's constant.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Operating temperature \( T = 1050 \, K \)
- Wien's constant \( b = 0.288 \, \text{cm K} = 0.288 \times 10^{-2} \, \text{m K} \)

2. **Use Wien's Displacement Law:**
According to the law:
\[ \lambda_{max} \cdot T = b \]

3. **Rearrange the Equation to Solve for \( \lambda_{max} \):**
\[ \lambda_{max} = \frac{b}{T} \]

4. **Substitute the Values:**
\[ \lambda_{max} = \frac{0.288 \times 10^{-2} \, \text{m K}}{1050 \, K} \]

5. **Calculate \( \lambda_{max} \):**
\[ \lambda_{max} = \frac{0.288 \times 10^{-2}}{1050} \]
\[ \lambda_{max} = 2.742857 \times 10^{-6} \, \text{m} \]

6. **Round the Result:**
\[ \lambda_{max} \approx 2.743 \times 10^{-6} \, \text{m} \]

### Final Answer:
The wavelength at which the filament radiates maximum intensity is approximately \( 2.743 \times 10^{-6} \, \text{m} \) or \( 2743 \, \text{nm} \).